A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?
Answers
Answer: 1.64 *10^19 electrons
Explanation: In order to the explain this problem we have to consider the following:
The current= charge/time; so
as the electrons move in the tungsten wire we have:
0.526 C/s= N electrons per second* charge of electron=
N electrons/s= 0.526/1.6*10^-19= 3.28 *10^18 electrons/s
Then, during 5 seconds will pass:
3.28 *10^18 electrons/s*5 5s= 1.64 *10^19 electrons
Answer:
1.64 x 10^19 electrons
Explanation:
The current is defined as I=ΔQ/Δt where ∆Q is the amount of charge flowing past a point in the filament. This charge is comprised of electrons that each carry charge of e = 1.602 × 10^-19 C. So ΔQ=Ne=IΔt and the number of electrons flowing through the filament in 5 s is N=IΔte=(0.526 A)(5 s)1.602×10^−19 C=1.64×10^19 electrons.
Related Questions
A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.N·m(b) Find the angular acceleration of the airplane when it is inlevel flight.rad/s2(c) Find the linear acceleration of the airplane tangent to itsflight path.m/s2
A net force of 25.0 N causes an object to accelerate at 4.00 m/s2. What is the mass of the object?
An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.
A car is decelerating at the rate of 2 km/s square. If its initial speed is 66 km/s, how long will it take the car to come to a complete stop?
Answers
Answer:
σ = 0.8 N/m
Explanation:
Given that
L = 12 cm
We know that 1 m = 100 cm
L = 0.12 m
The force ,F= 0.096 N
Lets take surface tension = σ
We know that surface tension is given as
Therefore the surface tension σ will be 0.8 N/m .
σ = 0.8 N/m
Final answer:
The surface tension of the liquid in air is 0.8 N/m.
Explanation:
To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.
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Answers
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad
Answers
Final answer:
This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.
Explanation:
This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.
To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.
Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:
0.35*F2 = 3.25*F1 (1)
Because the net force should also be zero in equilibrium, we have:
F1 + F2 = 205.8 (2)
Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.
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Final answer:
To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a force of approximately 114N is required
Explanation:
The subject question is a classic example of Torque problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The weight of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.
This torque is calculated as Torque = r * F = 1.8m (distance from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.
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Answers
Answer:
The first graph is showing the constant acceleration (1 m/s)
Explanation:
The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4
The last graph is showing constant velocity therefore there is no acceleration (a = 0)
Answer:
A
Explanation:
on edge
b. down to the floor/ground.
c. to your left.
d. to your right.
Answers
Answer:
b. down to the floor/ground
Explanation:
The direction of the angular velocity can be easily found by the use of right hand rule. This rule states that when we curl the fingers of right hand in the direction of the rotation of the object, the direction of the thumb gives the direction of angular velocity. We apply this rule to our situation. Since, the door opens outward and the hinges are on right, it means that the rotation of door is clockwise. So, when we curl the fingers of right hand in clock wise direction, the thumb will point in downward direction. So, the direction of angular velocity will be down to the floor/ground.
The correct option is:
b. down to the floor/ground
Answers
Answer:
The electromagnetic force tends to push the protons apart. (Like forces repel).
Explanation:
Answer:its the electromagnetic force
Explanation:
Protons are positive so repel themselves